A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion.If the period of oscillation is 11.2 s,what is the spring constant of the bungee cord,assuming it has negligible mass compared to that of the jumper?
A)17.6 N/m
B)21.1 N/m
C)28.2 N/m

Question

Quizplus · Accepted Answer

The period of oscillation for a mass-spring system is given by:T = 2π√(m/k)where m is the mass and k is the spring constant.Solving for k, we get:k = m(4π²/T²)Plugging in the given values, we get:k = (56.0 kg)(4π²/11.2 s²) = 21.1 N/m

A 560 Kg Bungee Jumper Jumps Off a Bridge and Undergoes